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3 years ago

should I tell penny, potato chip and used napkin that I've made new friends and moved on. I don't want to make them sad. :(

Physics

2 answers:

MrRa [10]3 years ago

7 0

Answer: ok

Explanation:

damaskus [11]3 years ago

5 0

Answer:

yes

Explanation:

Telling your friends that you moved on is completely okay. :D It saves an entire toxic friendship.

Yikes For Penny, Potato Chip, and Used Napkin:

at least you'll find different people!

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Fiesta28 [93]

Answer:

packets of pen

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3 years ago

A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the wel

Jet001 [13]

Answer:

The well is 23.3 m

Explanation:

As the bucket is lifted out of the well, energy in the man is being transferred to the bucket as gravitational potential energy.

Work done against gravity = mass * height * acceleration due to to gravity

W = mgh

5 920 J = 25.9 kg * h * 9.8 m/s²

h = 23.3 m

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3 years ago

If you were to stand in the exact center of a rotating disc, you would only have what kind of

Dmitriy789 [7]

Answer:

Tangential speed or Rotational speed

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3 years ago

ASK YOUR TEACHER A baseball with a mass of 146 g is thrown horizontally with a speed of 40.6 m/s (91 mi/h) at a bat. The ball is

RideAnS [48]

Answer:

Explanation:

mass of the ball = 146 g = 146 / 1000 = 0.146 kg

initial speed of the ball = 40.6 m/s

final speed of the ball = - 45.1 m/s

time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s

impulse, Ft = change in momentum = mv - mu = m (v-u)

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4 0

3 years ago

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

$\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

R = Rydberg constant, $R=1.097\times 10^7\ m^{-1}$

$\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

Solving above equation we get the value of final n is,

$n_f=2.04$

$n_f=2$

So, it will relax in the n = 2. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers